[计数 DP]Atcoder AGC001 E. BBQ Hard

答案是求

∑i≠j(ai+aj+bi+bjai+aj)

这就相当于从平面内的 (−ai,−bi) 走到 (aj,bj) 的方案数

把点放到平面上，DP就好了

#include <cstdio>#include <iostream>#include <algorithm>using namespace std;const int N=8010,P=1e9+7;int n,a[200010],b[200010],fac[N],inv[N];int f[N][N];inline int &F(int x,int y){ return f[x+2005][y+2005]; }inline void Pre(){  fac[0]=1; for(int i=1;i<=8000;i++) fac[i]=1LL*fac[i-1]*i%P;  inv[1]=1; for(int i=2;i<=8000;i++) inv[i]=1LL*(P-P/i)*inv[P%i]%P;  inv[0]=1; for(int i=1;i<=8000;i++) inv[i]=1LL*inv[i]*inv[i-1]%P;}inline int C(int x,int y){  return 1LL*fac[x]*inv[y]%P*inv[x-y]%P;}int main(){  freopen("1.in","r",stdin);  freopen("1.out","w",stdout);  scanf("%d",&n); Pre();  for(int i=1;i<=n;i++)    scanf("%d%d",&a[i],&b[i]),F(-a[i],-b[i])++;  for(int i=-2000;i<=2000;i++)    for(int j=-2000;j<=2000;j++)      F(i,j)=((long long)F(i,j)+F(i-1,j)+F(i,j-1))%P;  int ans=0;  for(int i=1;i<=n;i++) ans=((long long)ans+F(a[i],b[i])-C(a[i]+a[i]+b[i]+b[i],b[i]+b[i]))%P;  ans=1LL*ans*(P+1>>1)%P;  printf("%d\n",(ans+P)%P);  return 0;}